3.16.55 \(\int \frac {24 e^{x/2}-x-12 e^x x+6 e^{x/2} x \log (16 x^2)}{x} \, dx\)

Optimal. Leaf size=26 \[ 15-x-12 \left (e^x-e^{x/2} \log \left (16 x^2\right )\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14, 2194, 2288} \begin {gather*} 12 e^{x/2} \log \left (16 x^2\right )-x-12 e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(24*E^(x/2) - x - 12*E^x*x + 6*E^(x/2)*x*Log[16*x^2])/x,x]

[Out]

-12*E^x - x + 12*E^(x/2)*Log[16*x^2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1-12 e^x+\frac {6 e^{x/2} \left (4+x \log \left (16 x^2\right )\right )}{x}\right ) \, dx\\ &=-x+6 \int \frac {e^{x/2} \left (4+x \log \left (16 x^2\right )\right )}{x} \, dx-12 \int e^x \, dx\\ &=-12 e^x-x+12 e^{x/2} \log \left (16 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 0.92 \begin {gather*} -12 e^x-x+12 e^{x/2} \log \left (16 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24*E^(x/2) - x - 12*E^x*x + 6*E^(x/2)*x*Log[16*x^2])/x,x]

[Out]

-12*E^x - x + 12*E^(x/2)*Log[16*x^2]

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fricas [A]  time = 0.68, size = 20, normalized size = 0.77 \begin {gather*} 12 \, e^{\left (\frac {1}{2} \, x\right )} \log \left (16 \, x^{2}\right ) - x - 12 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(1/2*x)*log(16*x^2)-12*exp(x)*x+24*exp(1/2*x)-x)/x,x, algorithm="fricas")

[Out]

12*e^(1/2*x)*log(16*x^2) - x - 12*e^x

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giac [A]  time = 0.15, size = 20, normalized size = 0.77 \begin {gather*} 12 \, e^{\left (\frac {1}{2} \, x\right )} \log \left (16 \, x^{2}\right ) - x - 12 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(1/2*x)*log(16*x^2)-12*exp(x)*x+24*exp(1/2*x)-x)/x,x, algorithm="giac")

[Out]

12*e^(1/2*x)*log(16*x^2) - x - 12*e^x

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maple [A]  time = 0.08, size = 25, normalized size = 0.96

method	result	size
norman	\(-x -12 \,{\mathrm e}^{x}+12 \,{\mathrm e}^{\frac {x}{2}} \ln \left (16 x^{2}\right )\)	\(25\)
default	\(-x +12 \left (\ln \left (16 x^{2}\right )-2 \ln \relax (x )\right ) {\mathrm e}^{\frac {x}{2}}+24 \ln \relax (x ) {\mathrm e}^{\frac {x}{2}}-12 \,{\mathrm e}^{x}\)	\(34\)
risch	\(24 \ln \relax (x ) {\mathrm e}^{\frac {x}{2}}-6 i {\mathrm e}^{\frac {x}{2}} \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+12 i {\mathrm e}^{\frac {x}{2}} \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-6 i {\mathrm e}^{\frac {x}{2}} \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+48 \,{\mathrm e}^{\frac {x}{2}} \ln \relax (2)-12 \,{\mathrm e}^{x}-x\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x*exp(1/2*x)*ln(16*x^2)-12*exp(x)*x+24*exp(1/2*x)-x)/x,x,method=_RETURNVERBOSE)

[Out]

-x-12*exp(1/2*x)^2+12*exp(1/2*x)*ln(16*x^2)

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maxima [A]  time = 0.64, size = 20, normalized size = 0.77 \begin {gather*} 12 \, e^{\left (\frac {1}{2} \, x\right )} \log \left (16 \, x^{2}\right ) - x - 12 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(1/2*x)*log(16*x^2)-12*exp(x)*x+24*exp(1/2*x)-x)/x,x, algorithm="maxima")

[Out]

12*e^(1/2*x)*log(16*x^2) - x - 12*e^x

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mupad [B]  time = 1.05, size = 20, normalized size = 0.77 \begin {gather*} 12\,{\mathrm {e}}^{x/2}\,\ln \left (16\,x^2\right )-12\,{\mathrm {e}}^x-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 24*exp(x/2) + 12*x*exp(x) - 6*x*exp(x/2)*log(16*x^2))/x,x)

[Out]

12*exp(x/2)*log(16*x^2) - 12*exp(x) - x

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sympy [A]  time = 0.32, size = 19, normalized size = 0.73 \begin {gather*} - x + 12 e^{\frac {x}{2}} \log {\left (16 x^{2} \right )} - 12 e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*x*exp(1/2*x)*ln(16*x**2)-12*exp(x)*x+24*exp(1/2*x)-x)/x,x)

[Out]

-x + 12*exp(x/2)*log(16*x**2) - 12*exp(x)

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